# Nonlinear equations and systems on the SAT

Canonical: https://duckyhelper.com/learn/sat-math/nonlinear-equations-and-systems/
Updated: 2026-10-01

Nonlinear equations include square roots, absolute values, variables in a denominator, and systems where a line meets a parabola. Isolate the tricky part, then undo it: square both sides, split into two cases, or multiply by the denominator. Always check every answer in the original equation, because squaring and clearing denominators can create extraneous solutions that do not really work.

## The key idea

**How to undo each kind**

| Equation has | Move | Watch out for |
| --- | --- | --- |
| A square root | Isolate the root, then square both sides | Answers that make the root equal a negative number |
| An absolute value | Isolate it, then split into \(+\) and \(-\) cases | \(\|A\| = \) a negative number has no solution |
| x in a denominator | Multiply every term by the denominator | Answers that make a denominator 0 |
| A line and a parabola | Set the y's equal, solve the quadratic | 0, 1 or 2 intersection points |

Squaring is not reversible: \(3^2\) and \((-3)^2\) are both 9. So after squaring you may pick up an answer that fit the squared equation but not the original. That is an extraneous solution, and the SAT loves to offer it as a choice.

For a line-parabola system, substitution leads to a quadratic. Its discriminant counts the intersections:

$$
b^2 - 4ac > 0: \text{ two points} \qquad b^2 - 4ac = 0: \text{ one point} \qquad b^2 - 4ac < 0: \text{ none}
$$

## Worked examples

**Example 1: a radical equation with an extraneous solution**

Problem: Solve \(\sqrt{x + 7} = x - 5\).

1. The root is already alone. Square both sides.

   $$
   x + 7 = x^2 - 10x + 25
   $$
2. Move everything to one side.

   $$
   x^2 - 11x + 18 = 0
   $$
3. Factor.

   $$
   (x - 9)(x - 2) = 0
   $$
4. Check \(x = 9\): \(\sqrt{16} = 4\) and \(9 - 5 = 4\). It works.

   $$
   \sqrt{9 + 7} = 9 - 5
   $$
5. Check \(x = 2\): \(\sqrt{9} = 3\), but \(2 - 5 = -3\). It fails, so 2 is extraneous.

Answer: \(x = 9\)

**Example 2: absolute value**

Problem: Solve \(|2x - 3| = 7\).

1. The inside is either 7 or \(-7\). First case:

   $$
   2x - 3 = 7
   $$
2. Second case:

   $$
   2x - 3 = -7
   $$
3. Solve each: \(2x = 10\) gives \(x = 5\), and \(2x = -4\) gives \(x = -2\). Both check: \(|7| = 7\) and \(|-7| = 7\).

Answer: \(x = 5\) or \(x = -2\)

**Example 3: a rational equation with no solution**

Problem: Solve \(\frac{x}{x - 4} + 2 = \frac{4}{x - 4}\).

1. Multiply every term by \(x - 4\), including the 2.

   $$
   x + 2(x - 4) = 4
   $$
2. Simplify.

   $$
   3x - 8 = 4
   $$
3. Solve.

   $$
   x = 4
   $$
4. But \(x = 4\) makes the denominator \(x - 4\) equal 0, so it is not allowed. It is extraneous, and there is nothing else.

Answer: No solution

**Example 4 (SAT-hard): a line tangent to a parabola**

Problem: The system \(y = x^2 - 4x + 1\) and \(y = 2x + k\) has exactly one solution. What is the value of k?

1. Set the two expressions for y equal and move everything to one side.

   $$
   x^2 - 6x + (1 - k) = 0
   $$
2. Exactly one solution means the discriminant is 0.

   $$
   36 - 4(1 - k) = 0
   $$
3. Divide by 4: \(9 - (1 - k) = 0\), so \(8 + k = 0\).

   $$
   k = -8
   $$
4. Check: \(x^2 - 6x + 9 = (x - 3)^2\), so the line touches the parabola only at \(x = 3\), the point \((3, -2)\).

Answer: \(k = -8\)

## Common mistakes

- **Squaring before isolating the root.** Squaring \(\sqrt{x} + 2 = 5\) as is creates a messy middle term. Fix: subtract 2 first, then square.
- **Squaring \(x - 5\) as \(x^2 - 25\).** \((x - 5)^2 = x^2 - 10x + 25\). Fix: write it as \((x - 5)(x - 5)\).
- **Keeping both answers without checking.** Squaring can add a fake answer. Fix: plug each answer into the original equation.
- **Solving only the positive case.** \(|2x - 3| = 7\) has two cases. Fix: always write both \(= 7\) and \(= -7\).
- **Forgetting that a denominator cannot be 0.** If your answer makes any denominator 0, throw it out.

## Quick method

> **Tip: Desmos ignores extraneous answers**
>
> Graph each side as its own function, like \(y = \sqrt{x + 7}\) and \(y = x - 5\). Desmos only draws real values, so the graphs cross only at the true solution, \(x = 9\). It is a fast, honest way to catch extraneous answers. See [Desmos on the SAT](https://duckyhelper.com/learn/sat-math/desmos-calculator/).

## Practice

**5 SAT-style questions**

1. What is the solution to \(\sqrt{2x + 1} = 5\)?
   A. \(2\)
   B. \(12\)
   C. \(13\)
   D. \(24\)

   Answer: \(12\). Square both sides: \(2x + 1 = 25\), so \(2x = 24\) and \(x = 12\). Check: \(\sqrt{25} = 5\). 2 comes from \(2x + 1 = 5\), which forgets to square the 5. 13 adds the 1 instead of subtracting it, and 24 forgets to divide by 2.

2. What are all solutions to \(|x + 4| = 9\)?
   A. \(5\) and \(-13\)
   B. \(5\) only
   C. \(-5\) and \(13\)
   D. \(5\) and \(13\)

   Answer: \(5\) and \(-13\). Either \(x + 4 = 9\), giving 5, or \(x + 4 = -9\), giving \(-13\). "5 only" forgets the negative case. \(-5\) and 13 solve \(|x - 4| = 9\) instead.

3. What are all solutions to \(\sqrt{x + 2} = x\)?
   A. \(-1\) and \(2\)
   B. \(2\) only
   C. \(-1\) only
   D. No solution

   Answer: \(2\) only. Squaring gives \(x^2 - x - 2 = 0\), so \(x = 2\) or \(x = -1\). Check \(-1\): \(\sqrt{1} = 1\), not \(-1\). It is extraneous, so only 2 works.

4. Which ordered pair is a solution to the system \(y = x^2 - 1\) and \(y = x + 1\)?
   A. \((1, 0)\)
   B. \((2, 3)\)
   C. \((-2, 3)\)
   D. \((0, 1)\)

   Answer: \((2, 3)\). Set \(x^2 - 1 = x + 1\): \(x^2 - x - 2 = 0\), so \(x = 2\) or \(x = -1\). The points are \((2, 3)\) and \((-1, 0)\). The other choices each fit at most one of the equations.

5. Student-produced response: what is the positive solution to \(\frac{12}{x} = x + 1\)?

   Answer: 3. Multiply by x: \(12 = x^2 + x\), so \(x^2 + x - 12 = 0\) and \((x + 4)(x - 3) = 0\). The solutions are \(-4\) and 3, and the positive one is 3. Check: \(\frac{12}{3} = 4 = 3 + 1\).

## Frequently asked questions

### What is an extraneous solution?

It is an answer that your algebra produced but that does not work in the original equation. It shows up after squaring both sides or multiplying by an expression with x. That is why every answer to a radical or rational equation must be checked.

### Can an absolute value equation have no solution?

Yes. An absolute value is never negative, so \(|x - 2| = -5\) has no solution. If the equation is \(|x - 2| + 8 = 3\), isolate the absolute value first: \(|x - 2| = -5\), and stop there.

### How many times can a line and a parabola meet?

Zero, one or two times. Substitute to get a quadratic and look at its discriminant: positive means two points, zero means the line just touches (tangent), and negative means they never meet.

## Sources

- [College Board: SAT Math, Advanced Math skills (nonlinear equations, systems in 2 variables)](https://satsuite.collegeboard.org/sat/whats-on-the-test/math/types/advanced), accessed 2026-10-01

## Related

- [Quadratics on the SAT](https://duckyhelper.com/learn/sat-math/quadratics/)
- [Systems of linear equations on the SAT](https://duckyhelper.com/learn/sat-math/systems-of-equations/)
- [How to solve absolute value equations and inequalities](https://duckyhelper.com/learn/algebra-1/absolute-value/)
- [Rational expressions and equations](https://duckyhelper.com/learn/algebra-2/rational-expressions/)
- [SAT Math study guides](https://duckyhelper.com/learn/sat-math/)

## Try asking Ducky

- "Why does x = 2 not work even though I solved it right?"
- "Do I really have to check every answer?"
- "Show me the line and the parabola touching in Desmos."

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